derivative of 1/(z^2+1), chain rule
derivative of 1/(z^2+1), chain rule

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Last updated at May 29, 2023 by Teachoo

Question 3 Find the derivative of f given by f (x) = tan–1 𝑥 assuming it exists. 𝑓 (𝑥)=〖𝑡𝑎𝑛〗^(−1) 𝑥 Let 𝒚= 〖𝒕𝒂𝒏〗^(−𝟏) 𝒙 tan〖𝑦=𝑥〗 𝒙=𝐭𝐚𝐧〖𝒚 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (tan𝑦 ))/𝑑𝑥 1 = (𝑑 (tan𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = (𝑑 (tan𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = 〖𝐬𝐞𝐜〗^𝟐 𝒚 . 𝑑𝑦/𝑑𝑥 1 = (𝟏 + 𝒕𝒂𝒏𝟐𝒚) 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 1/(1 + 〖𝐭𝐚𝐧〗^𝟐𝒚 ) Putting 𝑡𝑎𝑛𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = 1/(1 + 𝒙^𝟐 ) Hence (𝒅(〖𝐭𝐚𝐧〗^(−𝟏)〖𝒙)〗)/𝒅𝒙 = 𝟏/(𝟏 + 𝒙^𝟐 ) As 𝑦 = 〖𝑡𝑎𝑛〗^(−1) 𝑥 So, 𝒕𝒂𝒏𝒚 = 𝒙 Derivative of 〖𝒄𝒐𝒔〗^(−𝟏) 𝒙 𝑓 (𝑥)=〖𝑐𝑜𝑠〗^(−1) 𝑥 Let 𝒚= 〖𝒄𝒐𝒔〗^(−𝟏) 𝒙 cos〖𝑦=𝑥〗 𝒙=𝐜𝐨𝐬〖𝒚 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (cos𝑦 ))/𝑑𝑥 1 = (𝑑 (cos𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = (𝑑 (cos𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = (−sin𝑦) 𝑑𝑦/𝑑𝑥 (−1)/sin𝑦 =𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (−1)/𝒔𝒊𝒏𝒚 𝑑𝑦/𝑑𝑥= (−1)/√(𝟏 − 〖𝒄𝒐𝒔〗^𝟐 𝒚) Putting 𝑐𝑜𝑠〖𝑦=𝑥〗 𝑑𝑦/𝑑𝑥= (−1)/√(1 − 𝒙^𝟐 ) Hence, (𝒅(〖𝒄𝒐𝒔〗^(−𝟏) 𝒙” ” ))/𝒅𝒙 = (−𝟏)/√(𝟏 − 𝒙^𝟐 ) “We know that” 〖𝑠𝑖𝑛〗^2 𝜃+〖𝑐𝑜𝑠〗^2 𝜃=1 〖𝑠𝑖𝑛〗^2 𝜃=1−〖𝑐𝑜𝑠〗^2 𝜃 𝒔𝒊𝒏𝜽=√(𝟏−〖𝒄𝒐𝒔〗^𝟐 𝜽) ” ” As 𝑦 = 〖𝑐𝑜𝑠〗^(−1) 𝑥 So, 𝒄𝒐𝒔𝒚 = 𝒙 Derivative of 〖𝒄𝒐𝒕〗^(−𝟏) 𝒙 𝑓 (𝑥)=〖𝑐𝑜𝑡〗^(−1) 𝑥 Let 𝒚= 〖𝒄𝒐𝒕〗^(−𝟏) 𝒙 cot〖𝑦=𝑥〗 𝒙=𝐜𝐨𝐭〖𝒚 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (cot𝑦 ))/𝑑𝑥 1 = (𝑑 (cot𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = (𝑑 (cot𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = −𝐜𝐨〖𝐬𝐞𝐜〗^𝟐 𝒚 . 𝑑𝑦/𝑑𝑥 1 = −(𝟏 +𝒄𝒐𝒕𝟐𝒚) 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (−1)/(1 + 〖𝐜𝐨𝐭〗^𝟐𝒚 ) Putting 𝑐𝑜𝑡𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = (−1)/(𝒙^𝟐 + 𝟏) Hence (𝒅(〖𝐜𝐨𝐭〗^(−𝟏)〖𝒙)〗)/𝒅𝒙 = (−𝟏)/(𝒙^𝟐 + 𝟏) (𝐴𝑠 〖 𝑐𝑜𝑠𝑒𝑐〗^2〖𝑦= 〖1+〗〖𝑐𝑜𝑡〗^2𝑦 〗) As 𝑦 = 〖𝑐𝑜𝑡〗^(−1) 𝑥 So, 𝒄𝒐𝒕𝒚 = 𝒙 Derivative of 〖𝒔𝒆𝒄〗^(−𝟏) 𝒙 𝑓 (𝑥)=〖𝑠𝑒𝑐〗^(−1) 𝑥 Let 𝒚= 〖𝒔𝒆𝒄〗^(−𝟏) 𝒙 sec〖𝑦=𝑥〗 𝒙=𝐬𝐞𝐜〖𝒚 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (sec𝑦 ))/𝑑𝑥 1 = (𝑑 (sec𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = (𝑑 (sec𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = 𝒕𝒂𝒏𝒚 .𝒔𝒆𝒄𝒚. 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 1/(𝒕𝒂𝒏𝒚 .〖 sec〗𝑦 ) 𝑑𝑦/𝑑𝑥 = 1/((√(〖𝐬𝐞𝐜〗^𝟐𝒚 − 𝟏)) .〖 sec〗𝑦 ) Putting value of 𝑠𝑒𝑐𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = 1/((√(𝑥^2 − 1 ) ) . 𝑥) 𝑑𝑦/𝑑𝑥 = 1/(𝑥 √(𝑥^2 − 1 ) ) Hence 𝒅(〖𝒔𝒆𝒄〗^(–𝟏) 𝒙)/𝒅𝒙 = 𝟏/(𝒙 √(𝒙^𝟐 − 𝟏 ) ) As tan2 θ = sec2 θ – 1, tan θ = √(“sec2 θ – 1” ) As 𝑦 = 〖𝑠𝑒𝑐〗^(−1) 𝑥 So, 𝒔𝒆𝒄𝒚 = 𝒙Derivative of 〖𝒄𝒐𝒔𝒆𝒄〗^(−𝟏) 𝒙 𝑓 (𝑥)=〖𝑐𝑜𝑠𝑒𝑐〗^(−1) 𝑥 Let 𝒚= 〖𝒄𝒐𝒔𝒆𝒄〗^(−𝟏) 𝒙 cosec〖𝑦=𝑥〗 𝒙=𝐜𝐨𝐬𝐞𝐜〖𝒚 〗 1 = (𝑑 (cosec𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = −cosec𝑦 .cot𝑦 . 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 1/(〖−cosec〗𝑦 .𝒄𝒐𝒕𝒚 ) 𝑑𝑦/𝑑𝑥 = 1/(〖−cosec〗𝑦 . √(〖𝐜𝐨𝒔𝒆𝒄〗^𝟐𝒚 − 𝟏)) Putting value of 𝑐𝑜𝑠𝑒𝑐𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = (−1)/(𝑥 √(𝑥^2 − 1 ) ) Hence 𝒅(〖𝒄𝒐𝒔𝒆𝒄〗^(–𝟏) 𝒙)/𝒅𝒙 = (−𝟏)/(𝒙 √(𝒙^𝟐 − 𝟏 ) ) As cot2 θ = cosec2 θ – 1, cot θ = √(“cosec2 θ – 1” ) Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (cosec𝑦 ))/𝑑𝑥 1 = (𝑑 (cosec𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 As cot2 θ = cosec2 θ – 1, cot θ = √(“cosec2 θ – 1” ) As 𝑦 = co〖𝑠𝑒𝑐〗^(−1) 𝑥 So, co𝒔𝒆𝒄𝒚 = 𝒙

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